To find which system of equations has the solution (-2, 1), we need to substitute x = -2 and y = 1 into each system and check if both equations hold true.
1) For 2x - y = -5 and x + 2y = 10:
2(-2) - 1 = -4 - 1 = -5 (holds true)
-2 + 2(1) = -2 + 2 = 0 ≠ 10 (does not hold true)
2) For y = -2x - 2 and 2x - 2y = 5:
1 = -2(-2) - 2 = 4 - 2 = 2 ≠ -2 (does not hold true)
2(-2) - 2(1) = -4 - 2 = -6 ≠ 5 (does not hold true)
3) For x - 4y = 8 and 3x + 10 = -2y:
-2 - 4(1) = -2 - 4 = -6 ≠ 8 (does not hold true)
3(-2) + 10 = -6 + 10 = 4 ≠ -2(1) = -2 (does not hold true)
4) For y = -3/2x - 2 and 2x - y = -5:
1 = -3/2(-2) - 2 = 3 + 2 = 5 ≠ -2 (does not hold true)
2(-2) - 1 = -4 - 1 = -5 (holds true)
Therefore, the system of equations 2x - y = -5 and y = -3/2x - 2 has the solution (-2, 1).
hich of the following systems of equations has the solution (-2, 1)?
This question requires you to show your work.
(1 point)
Responses
2x−y=−5
and x+2y=10
2 x minus y is equal to negative 5 and x plus 2 y is equal to 10
y=−2x−2
and 2x−2y=5
y is equal to negative 2 x minus 2 and 2 x minus 2 y is equal to 5
x−4y=8
and 3x+10=−2y
x minus 4 y is equal to 8 and 3 x plus 10 is equal to negative 2 y
y=−32x−2
and 2x−y=−5
1 answer
1 answer